Midpoint

Solution to Exercise 2.24.
The beginning is the half of every action.
— Greek Proverb

Exercise 2.24. Prove a segment has a unique midpoint.

Proof. Let be a segment. By the Ruler Placement Postulate there is a coordinate system (ruler) such that f(A) = 0 and f(B) > 0. Let We show that M is a midpoint of the segment. Since M is defined from the ruler f for the line through A and B, the points A, B, and M are collinear. Further,

Hence, M is a midpoint of the segment

We need to show that M is the unique midpoint of the segment. Suppose N is a midpoint of segment Then

|f(A) – f(N)| = d(A, N) = d(N, B) = |f(N) – f(B)|.

That is, |f(N)| = |f(N) – f(B)|. One of three cases must be true: or

Case 1. Assume Then |f(N)| = |f(N) – f(B)| implies –f(N) = –(f(N) – f(B)). Hence, f(B) = 0, but this contradicts that f(B) > 0.

Case 2. Assume Then |f(N)| = |f(N) – f(B)| implies f(N) = f(N) – f(B). Hence, f(B) = 0, but this contradicts that f(B) > 0.

Case 3. Assume Then |f(N)| = |f(N) – f(B)| implies f(N) = –(f(N) – f(B)). Hence Thus

Therefore, the point M is the unique midpoint of segment

Solutions for Chapter 2